1、SolutionsChemistry I2 Chapters 7 & 8SolutionsChemistry I2 ChapteSome DefinitionsA solution is a homogeneous mixture of 2 or more substances . One constituent is usually regarded as the SOLVENT and the others as SOLUTES.Some DefinitionsA solution is Parts of a SolutionSOLUTE the part of a solution th

2、at is being dissolved (usually the lesser amount)SOLVENT the part of a solution that dissolves the solute (usually the greater amount)Solute + Solvent = SolutionSoluteSolventExamplesolidsolidMetal alloyssolidliquidKool aidliquidliquidAlcoholic drinksgasliquidPepsigasgasairParts of a SolutionSOLUTE t

3、hDefinitionsSolutions can be classified as saturated or unsaturated.A saturated solution contains the maximum quantity of solute that dissolves at that temperature.An unsaturated solution contains less than the maximum amount of solute that can dissolve at a particular temperatureDefinitionsSolution

4、s can be clDefinitionsSUPERSATURATED SOLUTIONS contain more solute than is possible to be dissolvedSupersaturated solutions are unstable. The supersaturation is only temporary, and usually accomplished in one of two ways:Warm the solvent so that it will dissolve more, then cool the solution Evaporat

5、e some of the solvent carefully so that the solute does not solidify and come out of solution.DefinitionsSUPERSATURATED SOLUMiscibilityMiscible: compounds that dissolve readily in each other in any proportion.Ex: water and alcohol.Immiscible: Liquids that do not readily dissolve in each other Ex: oi

6、l and waterMiscibilityMiscible: compoundSolubilitySolubility is relative, most substances will dissolve by extremely small amounts.Soluble: the solute will dissolve in a particular solvent greater than 1 g / 100 mLInsoluble: the solute will dissolve only 0.1 g / 100 mL, or not dissolve in the solven

7、tSolubilitySolubility is relatiIONIC COMPOUNDSCompounds in Aqueous SolutionMany reactions involve ionic compounds, especially reactions in water aqueous solutions.KMnO4 in waterK+(aq) + MnO4-(aq)IONIC COMPOUNDSCompounds in AHow do we know ions are present in aqueous solutions?The solutions _They are

8、 called ELECTROLYTESHCl, MgCl2, and NaCl are strong electrolytes. They dissociate completely (or nearly so) into ions.Aqueous SolutionsHow do we know ions are presenAqueous SolutionsSome compounds dissolve in water but do not conduct electricity. They are called nonelectrolytes.Examples include:suga

9、rethanolethylene glycolAqueous SolutionsSome compoundConcentration of SoluteThe amount of solute in a solution is given by its concentration.Molarity(M)=moles soluteliters of solutionConcentration of SoluteThe amo1.0 L of water was used to make 1.0 L of solution. Notice the water left over.1.0 L of

10、water was used to makPROBLEM: Dissolve 5.00 g of NiCl26 H2O in enough water to make 250 mL of solution. Calculate the Molarity.Step 1: Calculate moles of NiCl26H2OStep 2: Calculate MolarityNiCl26 H2O = 0.0841 MPROBLEM: Dissolve 5.00 g of NStep 1: Change mL to L.250 mL * 1L/1000mL = 0.250 LStep 2: Ca

11、lculate.Moles = (0.0500 mol/L) (0.250 L) = 0.0125 molesStep 3: Convert moles to grams.(0.0125 mol)(90.00 g/mol) = 1.13 gUSING MOLARITYmoles = MVWhat mass of oxalic acid, H2C2O4, isrequired to make 250. mL of a 0.0500 Msolution?Step 1: Change mL to L.USING Practice For homework:Pg 242 # 1-6, 9Pg 268

12、# 19 & 20 (Read pg 266-268)Practice For homework:Pg 242 #Solutions (continued)Day 2Mrs. KaySolutions (continued)Day 2Learning CheckHow many grams of NaOH are required to prepare 400. mL of 3.0 M NaOH solution?1)12 g2)48 g3) 300 gLearning CheckHow many grams An IDEAL SOLUTION is one where the propert

13、ies depend only on the concentration of solute.Need conc. units to tell us the number of solute particles per solvent particle.The unit “molarity” does not do this!Concentration UnitsAn IDEAL SOLUTION is one whereTwo Other Concentration Unitsgrams solutegrams solutionMOLALITY, m% by mass =% by mass

14、m of solution= mol solutekilograms solventTwo Other Concentration UnitsgCalculating ConcentrationsDissolve 62.1 g (1.00 mol) of ethylene glycol in 250. g of H2O. Calculate molality and % by mass of ethylene glycol. Calculating ConcentrationsDissCalculating ConcentrationsCalculate molality Dissolve 6

15、2.1 g (1.00 mol) of ethylene glycol in 250. g of H2O. Calculate m & % of ethylene glycol (by mass).Calculate weight %Calculating ConcentrationsCalcLearning CheckA solution contains 15 g Na2CO3 and 235 g of H2O? What is the mass % of the solution? 1) 15% Na2CO32) 6.4% Na2CO33) 6.0% Na2CO3Learning Che

16、ckA solution contUsing mass %How many grams of NaCl are needed to prepare 250 g of a 10.0% (by mass) NaCl solution? Using mass %How many gramTry this molality problem25.0 g of NaCl is dissolved in 5000. mL of water. Find the molality (m) of the resulting solution.m = mol solute / kg solvent25 g NaCl

17、 1 mol NaCl 58.5 g NaCl= 0.427 mol NaClSince the density of water is 1 g/mL, 5000 mL = 5000 g, which is 5 kg0.427 mol NaCl 5 kg water= 0.0854 molal salt waterTry this molality problem25.0 Preparing SolutionsWeigh out a solid solute and dissolve in a given quantity of solvent.Dilute a concentrated so

18、lution to give one that is less concentrated.Preparing SolutionsWeigh out aPractice for homework:Read pg 255-263Do Pg 261 #5-8 (%m/m)Do Pg 268 # 19-20 (molarity)Practice for homework:Read pg Changing Concentration:Our calculations for this unit are based on solutions being carefully made with known

19、concentrations.Two ways to have a known concentration:Dissolving a measured mass of pure solute in a certain volume of solution. (We practiced this as molarity questions)Diluting a solution of known concentration to the new concentration youre trying to achieve.Changing Concentration:Our calDiluting

20、 a Standard solution:Use the formula:C1v1 = C2v2You will be given 3 of the 4 variables and need to solve for the unknown.C1 = the original concentrated solutionv1= the amount of original concentrated solution used.C2= the new diluted concentration of solutionv2= the new volume of diluted solutionDil

21、uting a Standard solution:UPractice:I have 2.0 L of 0.10 M of sulfuric acid. This usually sold as an 18 M concentrated solution. How much of the concentrated solution do I need to make my new solution?YOU KNOW: C1=18 M; C2=0.10 M; v2= 2.0 LSo, v1 = C2v2C1 v1 = (0.10 M)(2.0 L) (18 M) v1 = 0.011 L or

22、11 mLPractice:I have 2.0 L of 0.10 More Practice:Page 273 # 25-27Finish the Handout called “Molarity Review ” # 5-9More Practice:Page 273 # 25-27CHEM 11ASolution StoichiometryYou can solve for amounts of product made or reactants needed even though they are stated as mol/L rather than g or moles. You need to use the basics that you learned in stoichiometry from grade 11 chemistry and apply it to what youve learned about solutionsLets work through an